Решение:

a) $$(3\frac{7}{30}-1\frac{5}{12}):18\frac{1}{6}=(3\frac{14}{60}-1\frac{25}{60}):18\frac{1}{6}=(2\frac{74}{60}-1\frac{25}{60}):18\frac{1}{6}=1\frac{49}{60}:18\frac{1}{6}=\frac{109}{60}:\frac{109}{6}=\frac{109}{60}\cdot\frac{6}{109}=\frac{1}{10}=0,1.$$ Ответ: 0,1

б) $$(1\frac{1}{2}+2\frac{2}{3}):1\frac{2}{3}=(1\frac{3}{6}+2\frac{4}{6}):1\frac{2}{3}=3\frac{7}{6}:1\frac{2}{3}=4\frac{1}{6}:1\frac{2}{3}=\frac{25}{6}:\frac{5}{3}=\frac{25}{6}\cdot\frac{3}{5}=\frac{5}{2}\cdot\frac{1}{1}=\frac{5}{2}=2,5.$$ Ответ: 2,5

в) $$(\frac{11}{18}-1\frac{7}{12})\cdot(2\frac{1}{6}+\frac{7}{30})=(\frac{22}{36}-1\frac{21}{36})\cdot(2\frac{5}{30}+\frac{7}{30})=(\frac{22}{36}-\frac{57}{36})\cdot(2\frac{12}{30})=(\frac{-35}{36})\cdot(\frac{72}{30})=(\frac{-35}{36})\cdot(\frac{12}{5})=(\frac{-7}{3})\cdot(\frac{1}{1})=-\frac{7}{3}=-2\frac{1}{3}.$$

г) $$(3\frac{2}{5}-5)\cdot(\frac{31}{48}+\frac{7}{24})=(\frac{17}{5}-5)\cdot(\frac{31}{48}+\frac{14}{48})=(\frac{17}{5}-\frac{25}{5})\cdot(\frac{45}{48})=\frac{-8}{5}\cdot\frac{15}{16}=\frac{-1}{1}\cdot\frac{3}{2}=-\frac{3}{2}=-1,5.$$ Ответ: -1,5